HDU 1856 More is better

最新推荐文章于 2019-06-29 14:23:40 发布

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本文介绍了一个寻找最大朋友圈数量的问题及解决方案。通过并查集算法，解决如何从一系列直接朋友关系中找出最多能保持联系的朋友数量。该问题适用于理解并查集的基本原理及其应用。

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More is better

Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 327680/102400 K (Java/Others)
Total Submission(s): 21423 Accepted Submission(s): 7805

Problem Description

Mr Wang wants some boys to help him with a project. Because the project is rather complex, the more boys come, the better it will be. Of course there are certain requirements.

Mr Wang selected a room big enough to hold the boys. The boy who are not been chosen has to leave the room immediately. There are 10000000 boys in the room numbered from 1 to 10000000 at the very beginning. After Mr Wang's selection any two of them who are still in this room should be friends (direct or indirect), or there is only one boy left. Given all the direct friend-pairs, you should decide the best way.

Input

The first line of the input contains an integer n (0 ≤ n ≤ 100 000) - the number of direct friend-pairs. The following n lines each contains a pair of numbers A and B separated by a single space that suggests A and B are direct friends. (A ≠ B, 1 ≤ A, B ≤ 10000000)

Output

The output in one line contains exactly one integer equals to the maximum number of boys Mr Wang may keep.

Sample Input

Sample Output

  
   4
2


   
    
     Hint
    
A and B are friends(direct or indirect), B and C are friends(direct or indirect), 
then A and C are also friends(indirect).

 In the first sample {1,2,5,6} is the result.
In the second sample {1,2},{3,4},{5,6},{7,8} are four kinds of answers.

Author

lxlcrystal@TJU

Source

HDU 2007 Programming Contest - Final

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判断节点最多的

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int pre[10000010],ran[10000010];
int flag;
int find(int x)
{
	if(x==pre[x])
	return x;
	return pre[x]=find(pre[x]);
}
void merge(int x,int y)
{
	int fx=find(x);
	int fy=find(y);
	if(fx!=fy)
	{
		ran[fy]+=ran[fx];
		pre[fx]=fy;
		if(flag<ran[fy])
		flag=ran[fy];
	 } 	
}
int main()
{
	int n,i,j,a,b;
	while(scanf("%d",&n)!=EOF)
	{
		flag=1;
		for(i=1;i<10000010;i++)
		{
			pre[i]=i;
			ran[i]=1;//初始化 
		}
		for(i=1;i<=n;i++)
		{
			scanf("%d%d",&a,&b);
			merge(a,b);
		}
		printf("%d\n",flag);
	}
}